The second order rate constant of oxygen uptake by hemoglobin (Hb) to form oxyge
ID: 869153 • Letter: T
Question
The second order rate constant of oxygen uptake by hemoglobin (Hb) to form oxygemoglobin (HbO2) is 2.1 x10^6 M^-1s^-1 at 37 degrees celsius. The simpified equation for this reaction is: Hb(aq) +O2(aq) -->HbO2(aq)a) what is the rate of formation of HbO2 if the concentrations of Hb and O2 in the blood at the lungs are 8.0 x10^-6 and 1.5x10^-6 M?
b) what would be the rate constant expression if the reaction were pseudo-first order with respect to O2? Determine the half-life under these conditions.
C) what is the equilibrium constant expression for this reaction? The second order rate constant of oxygen uptake by hemoglobin (Hb) to form oxygemoglobin (HbO2) is 2.1 x10^6 M^-1s^-1 at 37 degrees celsius. The simpified equation for this reaction is: Hb(aq) +O2(aq) -->HbO2(aq)
a) what is the rate of formation of HbO2 if the concentrations of Hb and O2 in the blood at the lungs are 8.0 x10^-6 and 1.5x10^-6 M?
b) what would be the rate constant expression if the reaction were pseudo-first order with respect to O2? Determine the half-life under these conditions.
C) what is the equilibrium constant expression for this reaction? Hb(aq) +O2(aq) -->HbO2(aq)
a) what is the rate of formation of HbO2 if the concentrations of Hb and O2 in the blood at the lungs are 8.0 x10^-6 and 1.5x10^-6 M?
b) what would be the rate constant expression if the reaction were pseudo-first order with respect to O2? Determine the half-life under these conditions.
C) what is the equilibrium constant expression for this reaction?
Explanation / Answer
a. Reaction rate = (dCHbO2 / dt ) = k [Hb][ O2] (second order)
= 2.1 x106 M-1s-1 x 8.0 x10-6 M x 1.5x10-6 M = 2.52 x 10-5 M.s-1
b. Assume [Hb]=const. Then reaction rate expression:
(dCHbO2 / dt ) = k